An iron rod of length 2 m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is

Question

An iron rod of length 2 m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is (A) 19.6 × 1020 N/m2 (B) 19.6 × 1018 N/m2 (C) 19.6 × 1010 N/m2 (D) 19.6 × 1015 N/m2

Tardigrade Admin · Accepted Answer

Y =( F / A /Δ l / l)=((250 × 9.8/50 × 10-6)/(0.5 × 10-3)2) =(250 × 9.8/50 × 10-6) × (2/0.5 × 10-3) ⇒ 19.6 × 1010 N / m 2

Q. An iron rod of length $2 m$ and cross-sectional area of $50 m m^{2}$ stretched by $0.5 mm$ , when a mass of $250 k g$ is hung from its lower end. Young’s modulus of iron rod is

$19.6 \times 1 0^{20} N / m^{2}$

$19.6 \times 1 0^{18} N / m^{2}$

$19.6 \times 1 0^{10} N / m^{2}$

$19.6 \times 1 0^{15} N / m^{2}$

$Y = \frac{F / A}{Δ l / l} = \frac{\frac{250 \times 9.8}{50 \times 1 0 ^{- 6}}}{\frac{0.5 \times 1 0 ^{- 3}}{2}}$
$= \frac{250 \times 9.8}{50 \times 1 0 ^{- 6}} \times \frac{2}{0.5 \times 1 0 ^{- 3}}$
$\Rightarrow 19.6 \times 1 0^{10} N / m^{2}$

Q. An iron rod of length 2m and cross-sectional area of 50mm2 stretched by 0.5mm, when a mass of 250kg is hung from its lower end. Young’s modulus of iron rod is

19.6×1020N/m2

19.6×1018N/m2

19.6×1010N/m2

19.6×1015N/m2